Answer
True.
The required factor of $27+64{{n}^{3}}$ is $\left( 3+4n \right)\left( 9-12n+16{{n}^{2}} \right)$ .
Work Step by Step
Consider the expression.
$27+64{{n}^{3}}$
The given expression can be written as ${{\left( 3 \right)}^{3}}+{{\left( 4n \right)}^{3}}$.
Apply the formula of sum of cubes,
$\begin{align}
& 27+64{{n}^{3}}=\left( 3+4n \right)\left( {{3}^{2}}-3\cdot 4n+{{4}^{2}}{{n}^{2}} \right) \\
& =\left( 3+4n \right)\left( 9-12n+16{{n}^{2}} \right)
\end{align}$
Check:
$\begin{align}
& \left( 3+4n \right)\left( 9-12n+16{{n}^{2}} \right)=3\cdot 9-3\cdot 12n+3\cdot 16{{n}^{2}}+4n\cdot 9-4n\cdot 12n+4n\cdot 16{{n}^{2}} \\
& =27-36n+48{{n}^{2}}+36n-48{{n}^{2}}+64{{n}^{3}} \\
& =27-64{{n}^{3}}
\end{align}$
Thus, it is true.
